Let $f$ be a vector-valued function defined by $f(t)=(e^{-t},\cos(t))$. Find $f'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-e^{-t}-\sin(t)$ (Choice B) B $(-e^{-t},-\sin(t))$ (Choice C) C $(-\sin(t),-e^{-t})$ (Choice D) D $(-e^{-t},\sin(t))$
Answer: $f$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $f(t)=(e^{-t},\cos(t))$. Let's differentiate the first expression: $\dfrac{d}{dt}(e^{-t})=-e^{-t}$ Let's differentiate the second expression: $\dfrac{d}{dt}[\cos(t)]=-\sin(t)$ Now let's put everything together: $\begin{aligned} f'(t)&=\left(\dfrac{d}{dt}(e^{-t}),\dfrac{d}{dt}[\cos(t)]\right) \\\\ &=(-e^{-t},-\sin(t)) \end{aligned}$ In conclusion, $f'(t)=(-e^{-t},-\sin(t))$.